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3 years ago

Point h is between g and i. If gh = 2x2 + 7x, hi = 9x – 2, and gi = 38, find the length of gh, hi and the possible values of x.

1 answer:

4 0

2x²+7x+9x-2=38
2x²+16x-40=0
2(x²+8x-20)=0
2(x+10)(x-2)=0
x=-10 or x=2
if we let x=-10 we get negative lengths for some of the segments which is not allowed so discard it. for x=2,

gh=22, hi=16, and gi=38

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Homework 1:Pythagorean Theorem and it’s Converse Find The Value of X

Viefleur [7K]

The figure gives a dimension for the hypotenuse of 19

the height is given as 17, using the Pythagorean theorem we can find the base of the triangle:

19^2 = 17^2 + base^2

Simplify:

361 = 289 + base^2

Subtract 289 from both sides:

72 = base^2

Take the square root of both sides:

Base = 8.49 Round to 8.5

There are two identical triangles on each sides, so 8.5 x 2 = 17

The length of the two triangle bases is 17

The total base of the object is 31, so the length of the rectangular part in the middle would be 31 - 17 = 14

And X would be the same length as the bottom rectangle.

Therefore x = 14

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3 years ago

Read 2 more answers

2,17,82,257,626,1297 next one please ?

In-s [12.5K]

The easy thing to do is notice that 1^4 = 1, 2^4 = 16, 3^4 = 81, and so on, so the sequence follows the rule $n^4+1$ . The next number would then be fourth power of 7 plus 1, or 2402.

And the harder way: Denote the n-th term in this sequence by $a_n$ , and denote the given sequence by $\{a_n\}_{n\ge1}$ .

Let $b_n$ denote the n-th term in the sequence of forward differences of $\{a_n\}$ , defined by

$b_n=a_{n+1}-a_n$

for n ≥ 1. That is, $\{b_n\}$ is the sequence with

$b_1=a_2-a_1=17-2=15$

$b_2=a_3-a_2=82-17=65$

$b_3=a_4-a_3=175$

$b_4=a_5-a_4=369$

$b_5=a_6-a_5=671$

and so on.

Next, let $c_n$ denote the n-th term of the differences of $\{b_n\}$ , i.e. for n ≥ 1,

$c_n=b_{n+1}-b_n$

so that

$c_1=b_2-b_1=65-15=50$

$c_2=110$

$c_3=194$

$c_4=302$

etc.

Again: let $d_n$ denote the n-th difference of $\{c_n\}$ :

$d_n=c_{n+1}-c_n$

$d_1=c_2-c_1=60$

$d_2=84$

$d_3=108$

etc.

One more time: let $e_n$ denote the n-th difference of $\{d_n\}$ :

$e_n=d_{n+1}-d_n$

$e_1=d_2-d_1=24$

$e_2=24$

etc.

The fact that these last differences are constant is a good sign that $e_n=24$ for all n ≥ 1. Assuming this, we would see that $\{d_n\}$ is an arithmetic sequence given recursively by

$\begin{cases}d_1=60\\d_{n+1}=d_n+24&\text{for }n>1\end{cases}$

and we can easily find the explicit rule:

$d_2=d_1+24$

$d_3=d_2+24=d_1+24\cdot2$

$d_4=d_3+24=d_1+24\cdot3$

and so on, up to

$d_n=d_1+24(n-1)$

$d_n=24n+36$

Use the same strategy to find a closed form for $\{c_n\}$ , then for $\{b_n\}$ , and finally $\{a_n\}$ .

$\begin{cases}c_1=50\\c_{n+1}=c_n+24n+36&\text{for }n>1\end{cases}$

$c_2=c_1+24\cdot1+36$

$c_3=c_2+24\cdot2+36=c_1+24(1+2)+36\cdot2$

$c_4=c_3+24\cdot3+36=c_1+24(1+2+3)+36\cdot3$

and so on, up to

$c_n=c_1+24(1+2+3+\cdots+(n-1))+36(n-1)$

Recall the formula for the sum of consecutive integers:

$1+2+3+\cdots+n=\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2$

$\implies c_n=c_1+\dfrac{24(n-1)n}2+36(n-1)$

$\implies c_n=12n^2+24n+14$

$\begin{cases}b_1=15\\b_{n+1}=b_n+12n^2+24n+14&\text{for }n>1\end{cases}$

$b_2=b_1+12\cdot1^2+24\cdot1+14$

$b_3=b_2+12\cdot2^2+24\cdot2+14=b_1+12(1^2+2^2)+24(1+2)+14\cdot2$

$b_4=b_3+12\cdot3^2+24\cdot3+14=b_1+12(1^2+2^2+3^2)+24(1+2+3)+14\cdot3$

and so on, up to

$b_n=b_1+12(1^2+2^2+3^2+\cdots+(n-1)^2)+24(1+2+3+\cdots+(n-1))+14(n-1)$

Recall the formula for the sum of squares of consecutive integers:

$1^2+2^2+3^2+\cdots+n^2=\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6$

$\implies b_n=15+\dfrac{12(n-1)n(2(n-1)+1)}6+\dfrac{24(n-1)n}2+14(n-1)$

$\implies b_n=4n^3+6n^2+4n+1$

$\begin{cases}a_1=2\\a_{n+1}=a_n+4n^3+6n^2+4n+1&\text{for }n>1\end{cases}$

$a_2=a_1+4\cdot1^3+6\cdot1^2+4\cdot1+1$

$a_3=a_2+4(1^3+2^3)+6(1^2+2^2)+4(1+2)+1\cdot2$

$a_4=a_3+4(1^3+2^3+3^3)+6(1^2+2^2+3^2)+4(1+2+3)+1\cdot3$

$\implies a_n=a_1+4\displaystyle\sum_{k=1}^3k^3+6\sum_{k=1}^3k^2+4\sum_{k=1}^3k+\sum_{k=1}^{n-1}1$

$\displaystyle\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}4$

$\implies a_n=2+\dfrac{4(n-1)^2n^2}4+\dfrac{6(n-1)n(2n)}6+\dfrac{4(n-1)n}2+(n-1)$

$\implies a_n=n^4+1$

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3 years ago

in a grid of squares, Alice colored 3/4 of the squares blue. she colored 1/8 of the squares red. she colored the rest of the squ

Semmy [17]

Blue squares colored equals 6/8, add that to 1/8 (the red squares), this equals 7/8 which minus 8/8 is 1/8 which is the yellow colored squares

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PolarNik [594]

Answer:

one point

Step-by-step explanation:

A system of two linear equations will have one point in the solution set if the slopes of the lines are different.

When the equations are written in the same form, the ratio of x-coefficient to y-coefficient is related to the slope. It will be different if there is one solution.

ratio for first equation: 1/1 = 1
ratio for second equation: 1/-1 = -1

These lines have different slopes, so there is one solution to the system of equations.

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Additional comment

When the equations are in slope-intercept form with the y-coefficient equal to 1, the x-coefficient is the slope.

y = mx +b . . . . . slope = m

When the equations are in standard form (as in this problem), the ratio of x- to y-coefficient is the opposite of the slope.

ax +by = c . . . . . slope = -a/b

As long as the equations are in the same form, the slopes can be compared by comparing the ratios of coefficients.

If the slopes are the same, the lines may be either parallel (empty solution set) or coincident (infinite solution set). When the equations are in the same form with reduced coefficients, the lines will be coincident if they are the same equation.

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